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A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 weber/m². The coil carries a current of 2A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

  1. 0.20 Nm
  2. 0.24 Nm
  3. 0.12 Nm
  4. 0.15 Nm

Correct answer: 0.24 Nm

Solution

The torque on a current-carrying coil in a magnetic field is given by τ = nIBA sinθ, where n is the number of turns, I is the current, B is the magnetic field, A is the area of the coil, and θ is the angle between the plane of the coil and the magnetic field. Here, n = 50, I = 2 A, B = 0.2 T, A = 0.12 × 0.1 = 0.012 m², and θ = 30°. Substituting, τ = 50 × 2 × 0.2 × 0.012 × sin(30°) = 0.24 Nm.

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